Metamath Proof Explorer


Theorem sinaover2ne0

Description: If A in ( 0 , 2 _pi ) then sin ( A / 2 ) is not 0 . (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Assertion sinaover2ne0 ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 ≠ 0

Proof

Step Hyp Ref Expression
1 elioore ⊢ A ∈ 0 2 ⁢ π → A ∈ ℝ
2 1 recnd ⊢ A ∈ 0 2 ⁢ π → A ∈ ℂ
3 2cnd ⊢ A ∈ 0 2 ⁢ π → 2 ∈ ℂ
4 picn ⊢ π ∈ ℂ
5 4 a1i ⊢ A ∈ 0 2 ⁢ π → π ∈ ℂ
6 2ne0 ⊢ 2 ≠ 0
7 6 a1i ⊢ A ∈ 0 2 ⁢ π → 2 ≠ 0
8 pire ⊢ π ∈ ℝ
9 pipos ⊢ 0 < π
10 8 9 gt0ne0ii ⊢ π ≠ 0
11 10 a1i ⊢ A ∈ 0 2 ⁢ π → π ≠ 0
12 2 3 5 7 11 divdiv1d ⊢ A ∈ 0 2 ⁢ π → A 2 π = A 2 ⁢ π
13 0zd ⊢ A ∈ 0 2 ⁢ π → 0 ∈ ℤ
14 2re ⊢ 2 ∈ ℝ
15 14 8 remulcli ⊢ 2 ⁢ π ∈ ℝ
16 15 a1i ⊢ A ∈ 0 2 ⁢ π → 2 ⁢ π ∈ ℝ
17 0xr ⊢ 0 ∈ ℝ *
18 17 a1i ⊢ A ∈ 0 2 ⁢ π → 0 ∈ ℝ *
19 16 rexrd ⊢ A ∈ 0 2 ⁢ π → 2 ⁢ π ∈ ℝ *
20 id ⊢ A ∈ 0 2 ⁢ π → A ∈ 0 2 ⁢ π
21 ioogtlb ⊢ 0 ∈ ℝ * ∧ 2 ⁢ π ∈ ℝ * ∧ A ∈ 0 2 ⁢ π → 0 < A
22 18 19 20 21 syl3anc ⊢ A ∈ 0 2 ⁢ π → 0 < A
23 2pos ⊢ 0 < 2
24 14 8 23 9 mulgt0ii ⊢ 0 < 2 ⁢ π
25 24 a1i ⊢ A ∈ 0 2 ⁢ π → 0 < 2 ⁢ π
26 1 16 22 25 divgt0d ⊢ A ∈ 0 2 ⁢ π → 0 < A 2 ⁢ π
27 1rp ⊢ 1 ∈ ℝ +
28 27 a1i ⊢ A ∈ 0 2 ⁢ π → 1 ∈ ℝ +
29 16 25 elrpd ⊢ A ∈ 0 2 ⁢ π → 2 ⁢ π ∈ ℝ +
30 2 div1d ⊢ A ∈ 0 2 ⁢ π → A 1 = A
31 iooltub ⊢ 0 ∈ ℝ * ∧ 2 ⁢ π ∈ ℝ * ∧ A ∈ 0 2 ⁢ π → A < 2 ⁢ π
32 18 19 20 31 syl3anc ⊢ A ∈ 0 2 ⁢ π → A < 2 ⁢ π
33 30 32 eqbrtrd ⊢ A ∈ 0 2 ⁢ π → A 1 < 2 ⁢ π
34 1 28 29 33 ltdiv23d ⊢ A ∈ 0 2 ⁢ π → A 2 ⁢ π < 1
35 1e0p1 ⊢ 1 = 0 + 1
36 34 35 breqtrdi ⊢ A ∈ 0 2 ⁢ π → A 2 ⁢ π < 0 + 1
37 btwnnz ⊢ 0 ∈ ℤ ∧ 0 < A 2 ⁢ π ∧ A 2 ⁢ π < 0 + 1 → ¬ A 2 ⁢ π ∈ ℤ
38 13 26 36 37 syl3anc ⊢ A ∈ 0 2 ⁢ π → ¬ A 2 ⁢ π ∈ ℤ
39 12 38 eqneltrd ⊢ A ∈ 0 2 ⁢ π → ¬ A 2 π ∈ ℤ
40 2 halfcld ⊢ A ∈ 0 2 ⁢ π → A 2 ∈ ℂ
41 sineq0 ⊢ A 2 ∈ ℂ → sin ⁡ A 2 = 0 ↔ A 2 π ∈ ℤ
42 40 41 syl ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 = 0 ↔ A 2 π ∈ ℤ
43 39 42 mtbird ⊢ A ∈ 0 2 ⁢ π → ¬ sin ⁡ A 2 = 0
44 43 neqned ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 ≠ 0